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TIPD191: opa188

yang fan
Prodigy 60 points
Part Number: TIPD191
Other Parts Discussed in Thread: INA128, OPA188

Tool/software:

I am studying TIPD191 Instrumentation Amplifier with DC Rejection Reference Design recently, it's a inventive circuit. I have some questions about this circuit.

Is OPA188 and INA128 together forming a negative feedback loop to keep Vref consistent with offset of Vin? I think there is a loop between Vref and Vout. 

Below is integrator circuit of opa188 and simulations, a pole and a zero appear in curve of vin, frequency of pole is about 1.59Hz, which is created by R1 = 100k and C3 = 1uF,  frequency of zero is about 3.27kHz. I don't konw which compoents cause this zero, at frequency of 3.27kHz, impedence of C3 is about 50ohms, considering inverting amplifier  formed by R1 and C3, 20*log(50/100k)=-66dB, this value is same to  flatten area of high frequency above 3.27kHz of vin. So why does curve of vin become flat in high frequency?

  • 0
    TI__Mastermind 23728 points

    Yang,

    Integrator's can be tricky devices to simulate because they don't always develop a good DC operating point.  In fact, practical real world integrators can suffer from the same issue and drive into a supply rail.  For this reason sometimes, a modified integrator circuit is used called a practical integrator.( https://www.ti.com/lit/an/sboa275b/sboa275b.pdf ).  The practical integrator uses a very large feedback resistance to maintain a DC path for current to flow.  At most frequencies the integrator gain is dominated by the feedback capacitor impedance  and the input resistor impedance (C3 and R1 in your example).  At very low frequencies the gain will be set by the amplifier open loop gain or the ratio of the feedback resistor (let's call it Rf not shown in your diagram).

    Below, see a modified version of your circuit with Rf.  Note that the AC simulation shows the traditional expected 1/s transfer function for the integrator.  In your graph the flatlining of this function to a gain of 0dB is probably due to a DC operating point error.

    I hope this helps!

    Best regards, Art

  • 0 in reply to Art Kay
    Prodigy 60 points

    Hi, Art

    Thank you for your interpretation, I have viewed integrator's ciucuit in https://www.ti.com/lit/an/sboa275b/sboa275b.pdf  before, I just be curious about some strange  phenomenon that when I put off Rf and peform a simulation. For instance, can you explain reason of curve become bend toward up at high frequency in your graph? I suppose at high frequency C3 can be shorted hence Vos is same to Vout, but which determine this so-called high frequency? Besides there is a downward peak of Vos, how does it happen?

    Maybe I shouldn't ask these trivial questions, I just be curious and want to konw why.

    Best regards, Yang

  • 0 in reply to yang fan
    TI__Mastermind 23728 points

    Yang,

    When bode plots move up as in this case, the reason is normally the output impedance as Aol decreases.  This behavior is common on non-inverting amplifiers, and is especially seen when the open-loop output impedance is large or increases at higher frequency.  

    best regards,

    Art